Question

sinx-cosx-tanx=-1

Answer 1

\(\huge\color{red}{ \sf tan(x)= \frac{sin(x)}{cos(x)} }\)
so you would be getting

\(\huge\color{blue}{ \sf sin(x)-cos(x)- \frac{sin(x)}{cos(x)}=0 }\)




and \(\huge\color{red}{ \sf sin(x)=\sqrt{1-sin^2x} }\)

so so far you would have,

\(\huge\color{blue}{ \sf sin(x)-\sqrt{1-sin^2x} - \frac{sin(x)}{\sqrt{1-sin^2x} }=0 }\)

let sin(x)=a

\(\huge\color{blue}{ \sf a-\sqrt{1-a^2} - \frac{a}{\sqrt{1-a^2} }=0 }\)

Answer 2

can you solve this (looks abstruse, ik..)

Answer 3

or you could put everything in terms of cos and did it

Answer 4

\[\sin x-\cos x-\tan x+1=0
\]\[\cos x \left( \tan x-1 \right)-1\left( \tan x-1 \right)=0\]
\[\left( \tan x-1 \right)\left( \cos x-1 \right)=0\]
either \[\tan x=1=\tan n \frac{ \pi }{ 4 },x=n \frac{ \pi }{ 4 },~where ~n ~is~an~integer.\]
\[\cos x=1=\cos 2n \pi,x=2n \pi,~where~ n~ is~an~integer\]

Answer 5

correction
\[\tan x=\tan \left( n \pi+\frac{ \pi }{ 4 } \right),x=n \pi+\frac{ \pi }{ 4 }\]

Answer 6

alright that works, thank you.. Um, could you also have it in terms such as this.. sin x cos x - sin x - cos x^2 +cos x = 0
sin x (cos x - 1) - cos x (cos x - 1) = 0

Answer 7

it is correct also.

Answer 8

i got this by adding the -1 to both sides and then multiplying everything by cos x.

Answer 9

okie, thank you!!

Answer 10

just change tan x=sin x/cosx, and then multiply every term by cos x

Answer 11

yea, exactly. I think my problem was I forgot grouping and such

Answer 12

yw